What are the solutions to the equation 0=x^4-18x^2+32 ?

The solutions to the equation 0=x^4-18x^2+32 are x=4,x=-4,x=sqrt(2),x=-sqrt(2)

Find the zeros of 0=x^4-18x^2+32

The zeros of 0=x^4-18x^2+32 are x=4,x=-4,x=sqrt(2),x=-sqrt(2)

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0 = x^{4} - 18 x^{2} + 32

x = 4, x = - 4, x = 2, x = - 2

Solution steps

0 = x^{4} - 18 x^{2} + 32

Switch sides

x^{4} - 18 x^{2} + 32 = 0

Rewrite the equation with

u = x^{2}

and

u^{2} = x^{4}

u^{2} - 18 u + 32 = 0

Solve

u^{2} - 18 u + 32 = 0 : u = 16, u = 2

u = 16, u = 2

Substitute back

u = x^{2},

solve for

x

Solve

x^{2} = 16 : x = 4, x = - 4

Solve

x^{2} = 2 : x = 2, x = - 2

The solutions are

x = 4, x = - 4, x = 2, x = - 2

x (x - 5)^{3} - 14 (x - 5)^{2} = 0

4 x (x - 3)^{2} (2 x - 5) = 0

0 = x^{3} - 2 x^{2} - x + 2

x^{3} + x^{2} - 7 x + 15 = 0

a x^{3} - 2 x^{2} - 3 x = 0

What are the solutions to the equation 0=x^4-18x^2+32 ?
The solutions to the equation 0=x^4-18x^2+32 are x=4,x=-4,x=sqrt(2),x=-sqrt(2)
Find the zeros of 0=x^4-18x^2+32
The zeros of 0=x^4-18x^2+32 are x=4,x=-4,x=sqrt(2),x=-sqrt(2)